Here $u$ is harmonic.
Prove:
Let $B_1$ be the unit open ball in $\mathbb{R}^n$. $\bar{B_1}$ is its closure.
If $u$ is smooth in $\mathbb{R}^n-\bar{B_1}$ and continuous up to the boundary of $\mathbb{R}^n - B_1,$
On the boundary of $B_1$, $u$ always equals to zero.
There exists an epsilon greater than 0, such that $u(x) = O(|x|^{1-\epsilon})$ as $|x|$ goes to infinity.
Then $u = 0$ inside $ \mathbb{R}^n - B_1$
Here is the proof:
Consider the rotation matrix $O$ with $n$ columns and $n$ rows, $OO^T = O^TO= I$
we want to show $u(Ox) = 0$
Since $|\nabla_O u(Ox)| \leq^{\text{differentiating}} |x| |\nabla u(O x)| \leq^{\text{by gradient estimate}} |x| |x|^{-\epsilon}$
so as $x$ goes to infinity $|\nabla u(Ox)|$ goes to zero. We are done.
Here comes my confusion:
The gradient estimate usually comes the form:
$$\sup_{B_r(x_0)} |\nabla u| \leq \frac{c(n)}{r} \sup_{B_{2r}(x_0)} |u|$$
It is all about solid ball region. However, in our definition of harmonic function, the region we focus upon is the outside ball, and there might be some irregular things going inside the ball if we extend our harmonic function inside the ball.
But how can the author use the gradient estimate here? Do we have a gradient estimate for cut-out (I might use the wrong term here but you get the meaning) region?